average velocity=d/t
average velocity=(Vf+Vi)/2
then
2d/t=Vf+Vi
but a=(Vf-Vi)/t
or Vf=at+Vi
2d/t=at+Vi+Vi
2d/t^2=a+2Vi Now, if Vi is zero, you have it.
So the equation is only good when starting from zero.
I'm doing a lab and all i need to know is how to derive an equation
I'm accelrating over 5m on a flat surface and people timed it. I can average the time and now the time, distance, my weight and so forth
I took down the class notes and got
w = fd = (1/2)mv^2
ok and this
W=mad
and some how I'm suppose to derrive an equation getting this
a = (2d)/(t^2)
I don't know how to derrive that equation
2 answers
There is an equation that should have been in your class notes that says
d = (1/2) a t^2. That is what you need to derive a =(2d)/(t^2)
The equations that you wrote down in your class notes are correct but are not the ones you need for an easy derivation. But there is a way to do it with those equations
m a d = (1/2)m V^2
Cancel the m's and rearrange
a = V^2/(2d)
The average velocity during acceleration is Vav = V/2 = d/t
Therefore V^2 = 4 d^2/t^2, and
a = (4 d^2/t^2)/(2d) = 2d/t^2
d = (1/2) a t^2. That is what you need to derive a =(2d)/(t^2)
The equations that you wrote down in your class notes are correct but are not the ones you need for an easy derivation. But there is a way to do it with those equations
m a d = (1/2)m V^2
Cancel the m's and rearrange
a = V^2/(2d)
The average velocity during acceleration is Vav = V/2 = d/t
Therefore V^2 = 4 d^2/t^2, and
a = (4 d^2/t^2)/(2d) = 2d/t^2